I need help whith control pid

Hello: I am something desperate because I need help to control a plant with a PID. The idea is that it can find the values of Kp, Ki and Kd,
of the controller, since I have one but I do not have these values. the plant is the following one: G (s) @0/(1.5e-9s^2 +1) Thanks.
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On Wed, 30 May 2007 22:52:16 -0700, snipped-for-privacy@gmail.com wrote:

This almost has to be homework -- few real-world problems are so tidy, nor is one often called upon to control systems that resonate at 4kHz. Moreover, people with real-world problems usually start by saying something like "I'm trying to control this itty bitty piezo actuator with PID, and I'm having a problem...".
So tell us more! If it's homework go ahead and fess up -- most newsgroup populations will help with homework if you say so, and if you'll accept _help_ and not _answers_. If you're doing something real, please give us more details like your required performance, any interesting limitations (like hysteresis if it's a piezo actuator), and what you're trying to do it with (op-amps? DSP? FPGA?).
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Tim Wescott
Control systems and communications consulting
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snipped-for-privacy@gmail.com wrote:

There are an infinite number of solutions. Which one do you want?
Peter Nachtwey
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Peter Nachtwey wrote:

The third one from the end if the list?
Jerry
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Jerry, what is your solution? I have one that works for 3 real poles at -5.163978E4. I just arbitrarily picked 3 poles at 2 times the natural frequency.
Peter Nachtwey

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Peter Nachtwey wrote:
...

I'm sorry, Peter. I didn't even look at the problem. "Third from the end" of an infinite list was an attempt at humor.
Jerry
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+1) Thanks.

G(s) is undamped. Damp it!
Insert a damper in your system and your control works.
T^2 = 1.5e-9
G(s)_1 = 400/(T^2s^2 + 2*delta*T*s + 1)
E.g. delta = 1.1
See therefore http://home.arcor.de/janch/janch/_control/20070531-controldoc /
2*delta*T*s must be added!
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That isn't a PID. It isn't even closed loop control.
Peter Nachtwey
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On Fri, 01 Jun 2007 08:15:36 -0700, Peter Nachtwey wrote:
Newsbeitrag

Yet it's realistic, in an odd sort of way -- if you have a really nasty plant that you need to control, you should first ask if there's any chance of changing the plant.
The OP hasn't come back with sufficient detail to solve his problem. :(
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: On Fri, 01 Jun 2007 08:15:36 -0700, Peter Nachtwey wrote: : : Yet it's realistic, in an odd sort of way -- if you have a really nasty : plant that you need to control, you should first ask if there's any chance : of changing the plant.
That wouldn't help the OP with his tuning problem. : : The OP hasn't come back with sufficient detail to solve his problem. :(
The home work probably needed to be turned in by now. In any case the problem can be solved symbolically. Then one doesn't care if there is damping or not.
Peter Nachtwey
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Once I had to care about such a problem in practice.
It may be of some interest. I moved the set point according to the defined benchmark scheme for comparing with standard PID. The difference can be seen at page 3. As discussed before the controller range of 0.2...1 may not be exceeded.
If someone is interested: page 3 http://home.arcor.de/janch/janch/_control/20070531-controldoc /
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Newsbeitrag

You are still trying to make the PID look bad. This just proves you still can't calculate PID gains or know how to properly implement a PID. If have provided a few example in previous threads but you obviously keep ignoring them. This is also a non-solution to the wrong problem.
This one of an infinite number of solutions to julio's non-damped problem. ftp://ftp.deltacompsys.com/public/NG/Mathcad%20-%20t0p2%20simple%20OSC%20NG.pdf
In this example the closed loop transfer function has 3 poles so I chose a generic desired characteristic equation of ((s+a)^2+b^2) *(s +c). This means there is a real pole at -c and complex pole at -a-jb and -a+jb. I set a=c=3*natural frequency. I want the response to be faster than JCH's example just to show that the fault isn't with the PID, even if it wasn't realistically possible. I set b=0 because I wanted all three poles to be real. The PID gains are all calculated symbollically in terms of the system parameters and the desired characeristic equation. It is the desired response that julio was missing. By chosing where I want the poles I can get any closed loop response I want, in theory. The reality of sampling times and feedback resolution often limits how quick the response can be.
To get my example to work required an update time of 5 microseconds which is very fast.
Peter Nachtwey
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+ 2*delta*T*s + 1)
delta >= 1
Then the process can't oscillate for sure.
If you won't completely rebuild the system you have just to 'repair it mathematically' and add a damper like 2*1.1*T*s.
I was repairing a given system and you have completely build a new system! Our results cannot be compared.
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JCH wrote:

Yes, but that wasn't the problem. The problem is to find PID gains. You have a PID that you compare your over damped system to but you posted no gains or show how you got your results.

This works well if you are adding a few resistors to a circuit. It doesn't work on large mechanical systems. Think of the wasted energy. I have just added a PID as julio requested.

Again, you have modified the problem to fit you math. I have just added a PID to julio's system.

The PIDs can be compared. You can see your PID gets into position slowly and only after overshooting the set points. My complaint about your posts is that you make the PID look bad to prove what ever point you want to make. I have shown you how I calculate PID gains in previous threads but you ignore my examples and continue to make your PID examples look bad to what ever you are comparing them to. You also change the problem to fit the your math. Finally, you don't show your work or supply PID gains even though you had to calculate some for your comparison PID. You have not presented a solution to julio's problem.
Peter Nachtwey
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+ 2*delta*T*s + 1)

What if it would be a F-18 air fighter 'mechanical system' on all planes?
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On Fri, 01 Jun 2007 23:23:33 -0700, Peter Nachtwey wrote:

No it won't. But one of the saddest realities I eventually had to accept was that there are some closed-loop control problems that simply cannot be solved by a more-clever application of theory. Ultimately, every plant has it's limitations, and often the only way to get a system to come up to snuff is to change the plant, with more sensors, more actuators, or just plain old modifications.

True. The biggest reason I didn't want to reply substantively to his post is that without knowing what the expected range of good behavior of the plant is, you really can't go designing a controller to it. I know of at least four different ways of designing a controller for a highly resonant plant (not including changing the plant); each one applies to a different combination of plant characteristics and system requirements, and you cannot recommend any one of them without knowing a lot more about the plant than just a 2nd-order s-domain model.
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Tim Wescott
Control systems and communications consulting
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